You add 0.255 g of an orange, crystalline compound whose empirical formula is #C_10H_8Fe# to 11.12 g of benzene. The boiling point of the benzene rises from 80.10 C to 80.26 C. What are the molar mass and molecular formula of the compound?

Question

Noah Aldrich · Accepted Answer

The molecular formula is #"C"_20"H"_16"Fe"_2#, and the molar mass is 368.03 g.

The formula for boiling point elevation is #color(blue)(bar(ul(|color(white)(a/a) ΔT_"b" = K_"b"mcolor(white)(a/a)|)))" "# where #ΔT_"b"# = the increase in the boiling point #K_"b"# = the molal boiling point elevation constant of the solvent #m# = the molality of the solution We can rearrange the formula to get #m = (ΔT_"f")/K_"f"# In your problem, #ΔT_"f" = T_2 - T_1 = "80.26 °C - 80.10 °C" = "0.16 °C"# #K_"b" = "2.53 °C·kg·mol"^"-1"# ∴ #m = (0.16 color(red)(cancel(color(black)("°C"))))/(2.53 color(red)(cancel(color(black)("°C")))·"kg·mol"^"-1") = "0.0632 mol·kg"^"-1"# Now, #m = "0.0632 mol"/(1 color(red)(cancel(color(black)("kg")))) = "0.255 g"/("0.011 12" color(red)(cancel(color(black)("kg"))))# Divide both sides of the equilibrium by 0.0632. ∴ #"1 mol" = "0.255 g"/"0.011 12" × 1/0.0632 = "363 g"# The molar mass is 363 g/mol, so the molecular mass is 363 u. The empirical formula mass of #"C"_10"H"_8"Fe"# is 184.02 u. The molecular mass must be an integral multiple of the empirical formula mass. #"MM"/"EFM"= n# #n = "MM"/"EFM" = (363 color(red)(cancel(color(black)("u"))))/(184.02 color(red)(cancel(color(black)("u")))) = 1.97 ≈ 2# The molecular formula must be twice the empirical formula. #"MF" = ("C"_10"H"_8"Fe")_2 = "C"_20"H"_16"Fe"_2# Its molar mass is 368.03 g.

Kai Bowman · Answer

undefined The molar mass of the compound is 213.8 g/mol, and its molecular formula is $C_{16}H_{12}Fe$.